Problem: Let $g(x)=-x^3+4x^2-5x$. On which intervals is $g$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $1<x<\dfrac53$ only (Choice B) B $x>\dfrac53$ only (Choice C) C $x<1$ and $x>\dfrac53$ (Choice D) D $x<1$ only (Choice E) E The entire domain of $g$
We can analyze the intervals where $g$ is increasing/decreasing by looking for the intervals where its derivative $g'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $g$ is $g'(x)=(1-x)(3x-5)$. $g'(x)=0$ for $x=1,\dfrac53$. Since $g'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=1$ and $x=\dfrac53$. $g$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}1$ $0$ $1$ $2$ $3$ $ x<1$ $ 1<x<\frac{5}{3}$ $ \frac{5}{3}$ $ x>\frac{5}{3}$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $x<1$ $x=0$ $g'(0)=-5<0$ $g$ is decreasing $\searrow$ $1<x<\dfrac{5}{3}$ $x=\dfrac43$ $g'\left(\dfrac43\right)=\dfrac{1}{3}>0$ $g$ is increasing $\nearrow$ $x>\dfrac{5}{3}$ $x=2$ $g'(2)=-1<0$ $g$ is decreasing $\searrow$ In conclusion, $g$ is increasing over the interval $1<x<\dfrac53$ only.